What is the integral from 6 to infinity of xe^{-5x} ?

The integral from 6 to infinity of xe^{-5x} is (31)/(25e^{30)}

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integral from 6 to infinity of xe^{-5x}

\int_{6}^{\infty} x e^{- 5 x} d x

\frac{31}{25 e ^{30}}

Decimal Notation

1.16035 E - 13

Solution steps

\int_{6}^{\infty} x e^{- 5 x} d x

Compute the indefinite integral:

\int x e^{- 5 x} d x = \frac{1}{25} (- 5 e^{- 5 x} x - e^{- 5 x}) + C

Compute the boundaries:

\int_{6}^{\infty} x e^{- 5 x} d x = 0 - (- \frac{31}{25 e ^{30}})

= 0 - (- \frac{31}{25 e ^{30}})

Simplify

= \frac{31}{25 e ^{30}}

\frac{d}{d x} (2 ln (3 x^{2} - 1))

\frac{d}{d x} (- 19 x^{2} + 4 x + 22)

g (x) = x^{5} - 2 x

\int \frac{1}{x ^{2} - x + 6} d x

\int e^{- x y} d y

What is the integral from 6 to infinity of xe^{-5x} ?
The integral from 6 to infinity of xe^{-5x} is (31)/(25e^{30)}